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Question Number 164716 by cortano1 last updated on 21/Jan/22
   find minimum value of    f(x)=4sin 2x−5sin x−5cos x+6
$$\:\:\:{find}\:{minimum}\:{value}\:{of}\: \\ $$$$\:{f}\left({x}\right)=\mathrm{4sin}\:\mathrm{2}{x}−\mathrm{5sin}\:{x}−\mathrm{5cos}\:{x}+\mathrm{6} \\ $$
Answered by bobhans last updated on 21/Jan/22
 f(x)=4cos ((π/2)−2x)−5(√2) cos ((π/4)−x)+6   f(x)=4cos 2((π/4)−x)−5(√2) cos ((π/4)−x)+6   (π/4)−x=ϕ ⇒f(ϕ)=4cos 2ϕ−5(√2) cos ϕ+6  f(ϕ)=8cos^2 ϕ−5(√2) cos ϕ+2  f(ϕ) will be min when cos ϕ= ((5(√2))/(16))  min f(ϕ) = 8(((5(√2))/(16)))^2 −5(√2) (((5(√2))/(16)))+2=(7/(16))
$$\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{4cos}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{2x}\right)−\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{x}\right)+\mathrm{6} \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{4cos}\:\mathrm{2}\left(\frac{\pi}{\mathrm{4}}−\mathrm{x}\right)−\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{x}\right)+\mathrm{6} \\ $$$$\:\frac{\pi}{\mathrm{4}}−\mathrm{x}=\varphi\:\Rightarrow\mathrm{f}\left(\varphi\right)=\mathrm{4cos}\:\mathrm{2}\varphi−\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\varphi+\mathrm{6} \\ $$$$\mathrm{f}\left(\varphi\right)=\mathrm{8cos}\:^{\mathrm{2}} \varphi−\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\varphi+\mathrm{2} \\ $$$$\mathrm{f}\left(\varphi\right)\:\mathrm{will}\:\mathrm{be}\:\mathrm{min}\:\mathrm{when}\:\mathrm{cos}\:\varphi=\:\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{16}} \\ $$$$\mathrm{min}\:\mathrm{f}\left(\varphi\right)\:=\:\mathrm{8}\left(\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{16}}\right)^{\mathrm{2}} −\mathrm{5}\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{16}}\right)+\mathrm{2}=\frac{\mathrm{7}}{\mathrm{16}} \\ $$$$ \\ $$

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